BZOJ 3944

题目

戳这里

子问题：

$~~~~sum_{i=1}^{n}varphi(i)$

因为

$~~~~sum_{i|n}varphi(i)=n$

所以

$~~~~varphi(n)=n-sum_{i|n,i<n}varphi(i)$

辣么

$~~~~sum_{i=1}^{n}varphi(i)$

$=sum_{i=1}^{n}(i-sum_{d|i,d<i}varphi(d))$

$=frac{n(n+1)}{2}-sum_{i=2}^{n}sum_{d|i,d<i}varphi(d)$

$=frac{n(n+1)}{2}-sum_{i=2}^{n}sum_{d=1}^{lfloorfrac{n}{i} floor}varphi(d)$

同理
子问题：

$~~~~sum_{i=1}^{n}mu(i)$

因为

$~~~~sum_{i|n}mu(i)=[i==1]$

所以

$~~~~varphi(n)=[n==1]-sum_{i|n,i<n}varphi(i)$

辣么

$~~~~sum_{i=1}^{n}mu(i)$

$=sum_{i=1}^{n}([i==1]-sum_{d|i,d<i}mu(d))$

$=1-sum_{i=2}^{n}sum_{d|i,d<i}mu(d)$

$=1-sum_{i=2}^{n}sum_{d=1}^{lfloorfrac{n}{i} floor}mu(d)$

两个子问题都可以用杜教筛$O(n^{frac{2}{3}})$下求解

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<vector>
#include<iostream>
#include<algorithm>
 
#define maxn 5000000
 
using namespace std;
 
inline int getint()
{
    int num=0,flag=1;char c;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
    return num*flag;
}

map<long long,long long>Mu,Phi;
bool not_prime[maxn+5];
long long phi[maxn+5],mu[maxn+5],prime[maxn+5],cnt;
long long n;
 
inline void init()
{
    mu[1]=phi[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!not_prime[i])prime[++cnt]=i,mu[i]=-1,phi[i]=i-1;
        for(int j=1;j<=cnt&&i*prime[j]<=maxn;j++)
        {
            not_prime[i*prime[j]]=1;
            if(i%prime[j])mu[i*prime[j]]=-mu[i],phi[i*prime[j]]=phi[i]*phi[prime[j]];
            else{phi[i*prime[j]]=phi[i]*prime[j];break;}
        }
    }
    for(int i=1;i<=maxn;i++)phi[i]+=phi[i-1],mu[i]+=mu[i-1];
}
 
inline long long solvephi(long long x)
{
    if(x<=maxn)return phi[x];
    if(Phi.count(x))return Phi[x];
    long long Sum=1ll*x*(1ll*x+1)/2;
    for(long long i=2,j;i<=x;i=j+1)
        j=min(x/(x/i),x),Sum-=1ll*(j-i+1)*solvephi(x/i);
    return Phi[x]=Sum;
}
 
inline long long solvemu(long long x)
{
    if(x<=maxn)return mu[x];
    if(Mu.count(x))return Mu[x];
    long long Sum=1;
    for(long long i=2,j;i<=x;i=j+1)
        j=min(x/(x/i),x),Sum-=1ll*(j-i+1)*solvemu(x/i);
    return Mu[x]=Sum;
}
 
int main()
{
    init();
    int T=getint();
    while(T--)
    {
        n=getint();
        printf("%lld %lld
",solvephi(n),solvemu(n));
    }
}